If f⁢(a)=0fa0f(a)=0, then f⁢(x)=(x-a)⁢∑k=0n-2f(k+1)⁢(a)(k+1)!⁢(x-a)kfxxasuperscriptsubscriptk0n2superscriptfk1ak1superscriptxak\displaystyle f(x)=(x-a)\sum_{{k=0}}^{{n-2}}\frac{f^{{(k+1)}}(a)}{(k+1)!}(x-a)%
^{k}. Thus, f⁢(x)=(x-a)⁢g⁢(x)fxxagxf(x)=(x-a)g(x), where g⁢(x)gxg(x) is the polynomial ∑k=0n-2f(k+1)⁢(a)(k+1)!⁢(x-a)ksuperscriptsubscriptk0n2superscriptfk1ak1superscriptxak\displaystyle\sum_{{k=0}}^{{n-2}}\frac{f^{{(k+1)}}(a)}{(k+1)!}(x-a)^{k}. Hence, x-axax-a is a factor of f⁢(x)fxf(x).

Conversely, if x-axax-a is a factor of f⁢(x)fxf(x), then f⁢(x)=(x-a)⁢g⁢(x)fxxagxf(x)=(x-a)g(x) for some polynomial g⁢(x)gxg(x). Hence, f⁢(a)=(a-a)⁢g⁢(a)=0faaaga0f(a)=(a-a)g(a)=0.

It follows that x-axax-a is a factor of f⁢(x)fxf(x) if and only if f⁢(a)=0fa0f(a)=0.